Question

我正在尝试理解卡尔曼滤波器的最优性。根据维基百科：“从理论上可以知道，卡尔曼滤波器在a）模型与实际系统完全匹配的情况下是最优的，b）进入的噪声是白色的，c）噪声的协方差是完全已知的。” 。但这种最优性意味着什么，我该如何测试呢？

我在Kalman过滤器link上找到了学生daves示例并开始测试它。我通过改变卡尔曼滤波器估计的噪声参数来做到这一点我通过获取估计误差的均方根来对结果进行排序，并且当噪声方差与实际噪声参数匹配时，期望获得最佳结果。我错了。为什么是这样？

这是我在测试中使用的matlab代码（从Student Daves示例修改）

% written by StudentDave
%for licensing and usage questions
%email scienceguy5000 at gmail. com

%Bayesian Ninja tracking Quail using kalman filter

clear all
% define our meta-variables (i.e. how long and often we will sample)
duration = 10  %how long the Quail flies
dt = .1;  %The Ninja continuously looks for the birdy,
%but we'll assume he's just repeatedly sampling over time at a fixed interval

% Define update equations (Coefficent matrices): A physics based model for where we expect the Quail to be [state transition (state + velocity)] + [input control (acceleration)]
A = [1 dt; 0 1] ; % state transition matrix:  expected flight of the Quail (state prediction)
B = [dt^2/2; dt]; %input control matrix:  expected effect of the input accceleration on the state.
C = [1 0]; % measurement matrix: the expected measurement given the predicted state (likelihood)
%since we are only measuring position (too hard for the ninja to calculate speed), we set the velocity variable to
%zero.

% define main variables
u = 1.5; % define acceleration magnitude
Q= [0; 0]; %initized state--it has two components: [position; velocity] of the Quail
Q_estimate = Q;  %x_estimate of initial location estimation of where the Quail is (what we are updating)
QuailAccel_noise_mag =0.05; %process noise: the variability in how fast the Quail is speeding up (stdv of acceleration: meters/sec^2)
NinjaVision_noise_mag =10;  %measurement noise: How mask-blinded is the Ninja (stdv of location, in meters)
Ez = NinjaVision_noise_mag^2;% Ez convert the measurement noise (stdv) into covariance matrix
Ex = QuailAccel_noise_mag^2 * [dt^4/4 dt^3/2; dt^3/2 dt^2]; % Ex convert the process noise (stdv) into covariance matrix
P = Ex; % estimate of initial Quail position variance (covariance matrix)

% initize result variables
% Initialize for speed
Q_loc = []; % ACTUAL Quail flight path
vel = []; % ACTUAL Quail velocity
Q_loc_meas = []; % Quail path that the Ninja sees


% simulate what the Ninja sees over time
figure(2);clf
figure(1);clf
for t = 0 : dt: duration
    % Generate the Quail flight
    QuailAccel_noise = QuailAccel_noise_mag * [(dt^2/2)*randn; dt*randn];
    Q= A * Q+ B * u + QuailAccel_noise;
    % Generate what the Ninja sees
    NinjaVision_noise = NinjaVision_noise_mag * randn;
    y = C * Q+ NinjaVision_noise;
    Q_loc = [Q_loc; Q(1)];
    Q_loc_meas = [Q_loc_meas; y];
    vel = [vel; Q(2)];
    %iteratively plot what the ninja sees
    plot(0:dt:t, Q_loc, '-r.')
    plot(0:dt:t, Q_loc_meas, '-k.')
    axis([0 10 -30 80])
    hold on
end

%plot theoretical path of ninja that doesn't use kalman
plot(0:dt:t, smooth(Q_loc_meas), '-g.')

%plot velocity, just to show that it's constantly increasing, due to
%constant acceleration
%figure(2);
%plot(0:dt:t, vel, '-b.')


%% Do kalman filtering
Qa = 0.5*QuailAccel_noise_mag^2: 0.05*QuailAccel_noise_mag^2: 1.5*QuailAccel_noise_mag^2;
Ez3 = 0.5*Ez: 0.05*Ez: 1.5*Ez;
rms_tot = zeros(length(Qa), length(Ez3));
rms_tot2 = zeros(length(Qa), length(Ez3));
for i = 1:length(Qa)
    Ex2=Qa(i) * [dt^4/4 dt^3/2; dt^3/2 dt^2];
    for j=1:length(Ez3)
        Ez2=Ez3(j);
        %initize estimation variables
        Q_loc_estimate = []; %  Quail position estimate
        vel_estimate = []; % Quail velocity estimate
        Q= [0; 0]; % re-initized state
        P_estimate = P;
        P_mag_estimate = [];
        predic_state = [];
        predic_var = [];
        for t = 1:length(Q_loc)
            % Predict next state of the quail with the last state and predicted motion.
            Q_estimate = A * Q_estimate + B * u;
            predic_state = [predic_state; Q_estimate(1)] ;
            %predict next covariance
            P = A * P * A' + Ex2;
            predic_var = [predic_var; P] ;
            % predicted Ninja measurem  1ent covariance
            % Kalman Gain
            K = P*C'*inv(C*P*C'+Ez2);
            % Update the state estimate.
            Q_estimate = Q_estimate + K * (Q_loc_meas(t) - C * Q_estimate);
            % update covariance estimation.
            P =  (eye(2)-K*C)*P;
            %Store for plotting
            Q_loc_estimate = [Q_loc_estimate; Q_estimate(1)];
            vel_estimate = [vel_estimate; Q_estimate(2)];
            P_mag_estimate = [P_mag_estimate; P(2,2)];
        end
        rms_tot(i,j) = rms(Q_loc-Q_loc_estimate);
    end
end
close all;
figure;
surf(Qa, Ez3, rms_tot);
ylabel('measurement variance'); xlabel('model variance');

这是结果图片： link

为什么我错了？当估计的模型完全是真实模型时，为什么结果不是最好的？提前谢谢你：）

Answer 1

在上面提到的情况下，即1.模型是正确的2.噪声是白色高斯，已知方差卡尔曼滤波器是最优的，因为它是将获得最小均方误差（MMSE）并将达到的估计量CRB（Cramer Rao Bound）。

我会对数字结果发表评论，但上述代码不起作用，OP公布的结果链接已被破坏。卡尔曼滤波器是一种有效的迭代估计器，仅在MMSE意义上是最优的，即推导出估计器以最小化平方创新过程的预期值（假设为零均值白高斯过程）。在这种情况下，状态协方差矩阵是最低的（尽可能高）可以从无偏估计得到并且等于CRB。

卡尔曼滤波器的最优性

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